Question

One mole of $S{O}_3$ was placed in a vessel of $1$L capacity at a certain temperature when the following equilibrium was established.
$2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$
At equilibrium, $0.6$ moles of $S{O}_2$ were formed. The equilibrium constant $K_c$ for the reaction will be:

• A. $0.36$
• B. $0.45$
• C. $0.54$
• D. $0.675$

Answer 1

The correct option is D $0.675$

$2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$

Given initial concentration of $S{O}_3$= $1$ mole

$\therefore$ Concentration of $S{O}_3$ at equilibrium if $x$ is the amount of $S{O}_3$ reacted, is $1-2x$.

Given, at equilibrium, $\left[S{O}_2\right]=0.6$

$\Rightarrow 2x=0.6$

$x=0.3$

$\therefore \left[S{O}_3\right]=1-2x=1-0.6=0.4$

$\therefore \left[O_2\right]=x$

$\left[O_2\right]=0.3$

$\therefore K_c=\frac{\left[S{O}_2{]}^2\right[O_2]}{[S{O}_3{]}^2}$

$=\frac{\left(0.6{)}^2\right(0.3)}{(0.4{)}^2}$

$K_C=0.675$